Question

A line segment is bisected by a line with the equation `-7 y + 5 x = 1`. If one end of the line segment is at `(1 ,4 )`, where is the other end?

Answer 1

I get `(157/37 , -20/37 )`

Explanation:

`-7 y + 5x = 1`

I find it less confusing this way:

`5x - 7y = 1`

The perpendicular family is gotten by swapping the coefficients on `x` and `y`, negating one. The constant is gotten by plugging in the point `(1,4)` on the perpendicular:

`7x + 5y = 7(1) + 5(4) = 27`

We find the meet by multiplying the first by 5 and the second by 7:

`25 x - 35 y = 5`

`49 x + 35y = 7 cdot 27 = 189`

Adding,

`74 x = 194`

`x = 194/74 = 97/37`

`y = 1/5 (27 - 7(97/37)) = 64/37`

If we call our endpoint E and our meet M we get an informal equation for the other endpoint F that's

`F = M-(E-M)= M + (M-E) = 2M - E`

So our other endpoint is

`(2 (97/37) - 1, 2( 64/37) -4 ) = (157/37 , -20/37 )`

Check:

Let's see if we can get the grapher to graph it:

`-7 y + 5x = 1`

`(y-4)(157/37 -1)=(x-1)( -20/37 -4 )`

`( -7 y + 5x - 1) ( (y-4)(157/37 -1) - (x-1)( -20/37 -4 ) ) = 0`

graph{ ( -7 y + 5x - 1) ( (y-4)(157/37 -1) - (x-1)( -20/37 -4 ) ) = 0 [-7.83, 12.17, -2.44, 7.56]}

Looks pretty good.